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Let $V$ be a finite dimensional real vector space. I'm trying to develop an intuitive understanding of what it means for two ordered bases $\alpha = (u_1,\ldots,u_n)$ and $\beta = (v_1,\ldots,v_n)$ of $V$ to have the same orientation.

A popular definition is that $\alpha$ and $\beta$ have the same orientation if and only if the change of basis matrix from $\alpha$ to $\beta$ has a positive determinant. (For example, on p. 238 [section 21.1] of An Introduction to Manifolds by Tu, there is a statement, "We say that two ordered bases are equivalent if the change-of-basis matrix $A$ has positive determinant.")

However, this definition does not seem very intuitively clear to me. I can't see directly what this definition means in an intuitive way.

Here is an attempt at a more intuitive definition. Ordered bases $\alpha$ and $\beta$ have the same orientation if and only if there exist continuous functions $w_i:[0,1] \to V$ (for $i = 1,\ldots, n$) such that:

  1. The set $\{w_1(t),\ldots,w_n(t)\}$ is linearly independent for all $t \in [0,1]$.
  2. $w_i(0) = u_i$ for $i = 1,\ldots, n$.
  3. $w_i(1) = v_i$ for $i = 1,\ldots, n$.

Intuitively, this definition means that $\alpha$ can be "continuously deformed" into $\beta$ without passing through a degenerate configuration. I like this because it matches my intuition from $\mathbb R^3$: when I try to imagine deforming a right-handed basis into a left-handed basis, I always have to pass through a degenerate configuration. I can also visualize one and two dimensional examples.

I would like to show that this definition is equivalent to the more standard definition that I gave at the beginning. Here is an attempted proof. First of all, suppose that $\alpha$ and $\beta$ have the same orientation according to my proposed definition. Let $R(t)$ be the change of basis matrix from $\alpha$ to the ordered basis $(w_1(t),\ldots, w_n(t))$. Because $\det R(t)$ is a continuous function that is initially equal to $1$ and is never $0$, it must be true that $\det R(t) > 0$ for all $t \in [0,1]$. In particular, $\det R(1) > 0$. This shows that the change of basis matrix from $\alpha$ to $\beta$ has positive determinant.

Conversely, I need to show that if $\alpha$ and $\beta$ have the same orientation according to the standard definition (i.e., the change of basis matrix from $\alpha$ to $\beta$ has positive determinant), then $\alpha$ and $\beta$ have the same orientation according to my proposed definition.

My question: How do I show this?

I've considered some ideas based on using the SVD of the change of basis matrix, or the QR factorization, but I'm not sure they will work. (Also, I suspect a simpler approach should be available.)

Additional question: Please provide any comments that you think might be relevant, or tangentially relevant -- anything that you suspect might improve my understanding of this topic. Maybe a completely different way of looking at orientation would be more intuitive or enlightening.

eternalGoldenBraid
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  • "Standard definition"? From where or by whom? I never heard such a thing like "orientation of basis in a general basis of a vector space..." – DonAntonio Mar 23 '17 at 08:21
  • @DonAntonio For example, on p. 238 (section 21.1) of An Introduction to Manifolds by Tu, there is a statement, "We say that two ordered bases are equivalent if the change-of-basis matrix $A$ has positive determinant." I've used the phrase "have the same orientation" rather than "equivalent", which might be nonstandard. – eternalGoldenBraid Mar 23 '17 at 08:37
  • You could try to use the QR-decomposition of $R(1)$. If $R(1)$ has positive determinant, then there is an orthogonal $Q$ with positive determinant and an upper triangular $R$ with positive determinant such that $QR=R(1)$. Now you only need to find continuous and non-degenerate functions $q:[0,1]\rightarrow\mathbb{R}^{n\times n}$ with $q(0)=I_n$ and $q(1)=Q$ and $r:[0,1]\rightarrow\mathbb{R}^{n\times n}$ with $r(0)=I_n$ and $r(1)=R$ and you can set $w(t)=q(t)r(t)$. I must admit, I do not know how to construct $q$ right now, but maybe this is at least a starting point. – Reinhard Meier Mar 23 '17 at 08:45
  • @eternalGoldenBraid Oh, I know that, even with the non-standard name ...but you tagged your question simply "linear algebra" . No manifolds and etc. at all... – DonAntonio Mar 23 '17 at 08:47
  • @DonAntonio I added the "smooth-manifolds" tag, thanks. – eternalGoldenBraid Mar 23 '17 at 08:56

2 Answers2

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For the opposite direction, I have a solution which is certainly overkilled but short. Consider $\alpha, \beta$ two basis with the same orientation. We can assume $\alpha = (e_1, \dots, e_n)$. This means that $\beta$ has positive determinant. Let $B$ be the matrix of $\beta$, there is a matrix $C$ with $\exp(C) = B$ since $B$ has positive determinant. Now you can take $B_t = \exp(tC)$. We have $B_0 = \alpha$ and $B_1 = \beta$.

  • This is interesting. Is it a problem that we are working in a real vector space rather than in $\mathbb R^n$? I think this makes it unclear what the statement "$\beta$ has a positive determinant" means. – eternalGoldenBraid Mar 23 '17 at 08:43
  • No problem. I said "we can assume that $\alpha = (e_1, \dots, e_n)$ for use the surjectivity of $\exp : \mathfrak gl_n \to \rm{GL}_n^+(\mathbb R)$. Meaning of "positive determinant" : $\beta$ is a basis in $\mathbb R^n$ with the usual basis, so there is a matrix of base change from $\alpha$ to $\beta$ and I ask this matrix to have positive determinant. But of course any $\mathbb R$-vector space is isomorphic to some $\mathbb R^n$. –  Mar 23 '17 at 08:46
  • But a matrix with positive determinant still can have negative eigenvalues, in which case $C$ might not be real anymore, and it is not obvious that $\exp(tC)$ is always real in this case. – Reinhard Meier Mar 23 '17 at 08:49
  • Exponential of a real matrix is still real, as exponential of a real number is still real. –  Mar 23 '17 at 08:50
  • But $C$ is the logarithm, which is not necessarily real. – Reinhard Meier Mar 23 '17 at 08:51
  • You are right and there are matrix with positive determinant which cannot be written as $e^A$ with $A$ real. Sorry for the wrong solution. –  Mar 23 '17 at 08:53
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Let $B$ be the matrix that represents the change of the bases. We can use the QR-decomposition of $B$. If $B$ has positive determinant, then there is an orthogonal $Q$ with $\det(Q)=1$ and an upper triangular $R$ with positive elements on the diagonal such that $QR=B$. Now we only need to find continuous and non-degenerate functions $$ q:[0,1]\rightarrow\mathbb{R}^{n\times n} \;\;\mbox{with}\;\; q(0)=I_n,\;q(1)=Q \\ r:[0,1]\rightarrow\mathbb{R}^{n\times n} \;\;\mbox{with}\;\; r(0)=I_n, \; r(1)=R $$ and you can set $w(t)=q(t)r(t)$.

The construction of $r$ is trivial, we can use $$r(t) = (1-t)I_n + tR.$$ As the diagonal elements of $R$ are positive and the diagonal elements of $r(t)$ are convex combinations of the diagonal elements of $R$ and $I$, the diagonal elements of $r(t)$ will be positive, too.

For the construction of $q$, we recall that every orthogonal matrix $Q$ with $\det(Q)=1$ is a rotation, i.e. $Q=SD(\alpha_1,\ldots,\alpha_s)S^{-1}$ with

$$ D(\alpha_1,\ldots,\alpha_s) = \begin{pmatrix} \cos\alpha_1 & -\sin\alpha_1 & & & & & \\ \sin\alpha_1 & \cos\alpha_1 & & & & & \\ & & \cos\alpha_2 & -\sin\alpha_2 & & & \\ & & \sin\alpha_2 & \cos\alpha_2 & & & \\ & & & & \ddots & & \\ & & & & & \cos\alpha_s & -\sin\alpha_s \\ & & & & & \sin\alpha_s & \cos\alpha_s \end{pmatrix} $$ (for even $n$) or $$ D(\alpha_1,\ldots,\alpha_s) = \begin{pmatrix} \cos\alpha_1 & -\sin\alpha_1 & & & & & & \\ \sin\alpha_1 & \cos\alpha_1 & & & & & & \\ & & \cos\alpha_2 & -\sin\alpha_2 & & & & \\ & & \sin\alpha_2 & \cos\alpha_2 & & & & \\ & & & & \ddots & & &\\ & & & & & \cos\alpha_s & -\sin\alpha_s & \\ & & & & & \sin\alpha_s & \cos\alpha_s & \\ & & & & & & & 1 \end{pmatrix} $$ (for odd $n$).

With these definitions, we can set $$q(t) = SD(t\alpha_1,\ldots,t\alpha_s)S^{-1}.$$

Reinhard Meier
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